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Chapter 12: Hypothesis Tests – Large Samples
Summary Information
:
=150,
(number of successes) = 110,
= 0.02,
= 0.01,
⁄
= 2.33, and
= 0.66. Also,
√
=
5.8017.
Since the senior would like to establish that the proportion of teenagers in
his high school who were concerned about air quality is not equal to 66%,
the alternative hypothesis should reflect this belief.
Solution:
0.66
0.66
√
= (110 - 150
0.66)/5.8017 = 1.8960.
: For a significance level of
= 0.02, reject the null hypothesis if
= 1.8960 >
= 2.33 or
= 1.8960 < -
= -2.33. This is equivalent
to checking whether |
| = 1.8960 >
= 2.33.
Conclusion:
Since
= 1.8960 <
= 2.33, then the null hypothesis
will
not be rejected. That is, there is insufficient sample evidence to claim that
the proportion of teenagers who were concerned about air quality is not
equal to 66%, at the 2% level of significance.
Note:
This means that there is not a statistical significant difference between
the sample proportion of 0.7333 and the postulated proportion of 0.66.
Figure 12-7
displays the test statistic in relation to the rejection region.
Observe that the test statistic falls in the do not reject region.